No redshift in a freely falling frame

[PeterDonis]

I said that it was your method that was incorrect

The general method he used is correct, provided you adopt the particular coordinates and simultaneity convention he adopted. (The particular derivation he gave was for a special case, yes.) If you are saying those coordinates and that simultaneity convention are not the only possible ones, that's true; and in different coordinates you would use a different method for defining "relative clock rates". An observer on a satellite in low Earth orbit, for example, would see Earthbound clocks (clocks at rest on the Earth's surface and rotating with it) to be "running slow" if he used his own local inertial coordinates; but in the ECI frame his clocks would be "running slow" relative to Earthbound clocks (I believe that's right for low enough orbits--the GPS satellite orbits are very high, 4.2 Earth radii IIRC). However, the observed frequency shift for light traveling between the two observers would be independent of which coordinates you adopted.

 

[stevendaryl]

The general method he used is correct, provided you adopt the particular coordinates and simultaneity convention he adopted.

I don't see that his derivation works in the case of two observers on the surface of the Earth, one moving east and one moving west, with relative speed v. His method would give a redshift of zero (or at best, of the order of (v/c)²), which is not the correct answer. The correct answer should, in the limit as r_s→0 approach the Doppler shift formula.

(The particular derivation he gave was for a special case, yes.) If you are saying those coordinates and that simultaneity convention are not the only possible ones, that's true; and in different coordinates you would use a different method for defining "relative clock rates". An observer on a satellite in low Earth orbit, for example, would see Earthbound clocks (clocks at rest on the Earth's surface and rotating with it) to be "running slow" if he used his own local inertial coordinates; but in the ECI frame his clocks would be "running slow" relative to Earthbound clocks (I believe that's right for low enough orbits--the GPS satellite orbits are very high, 4.2 Earth radii IIRC). However, the observed frequency shift for light traveling between the two observers would be independent of which coordinates you adopted.

That's exactly my point: redshift is coordinate-independent, while "relative clock rate" is coordinate-dependent. The two are only equal in certain circumstances.

 

[GAsahi]

My mistake, but either way, it's incorrect in the case that I was talking about, namely two observers, at the equator, moving at relative speed v in opposite directions (east for one, west for the other). In that case
\frac{d \tau_1}{d \tau_2} = 1.
\frac{f_2}{f_1} = √((1-v/c)/(1+v/c))

First off, the formula \frac{f_2}{f_1} = \sqrt{((1-v/c)/(1+v/c))}
does not apply for circular motion. I know for a fact that the relativistic Doppler is more complicated for accelerated motion. Note the presence of the terms in \frac{v^2}{c^2} here.
More importantly, you do not get redshift as reflected in your formula, you get blueshift for half the circle (decreasing closing distance) and redshift only for the other half (increasing closing distance) so, on average, \frac{f_1}{f_2}=1 due to problem symmetry.
Lastly, we were discussing the correctness of my formulas/derivations for radial motion. Failing to prove your point , you quietly moved the goalposts to circular motion. But even then, you failed to acknowledge my repeated references to the Ashby paper that shows how this problem gets solved for the case of circular motion. The solution uses a different metric but the methodology is the same, all the effects (with the notable excption of the Sagnac effect) fall out the metric.

 

[PeterDonis]

I don't see that his derivation works in the case of two observers on the surface of the Earth, one moving east and one moving west, with relative speed v. His method would give a redshift of zero (or at best, of the order of (v/c)²), which is not the correct answer. The correct answer should, in the limit as r_s→0 approach the Doppler shift formula.

The correct answer for what? For the observed frequency shift of light sent between the two observers? Or for the relative clock rates of the two observers? You yourself pointed out that the two are not necessarily the same; this is a case where they are different.

A more general formula for the rate of a moving clock relative to coordinate time, generalizing the one that GAsahi derived, is this:

\frac{d\tau}{dt} = \sqrt{1 - \frac{2 G M}{c^{2} r}}\sqrt{1 - \frac{v^{2}}{c^{2}}}

which is similar to the formula in Ashby's paper, except that I have used Schwarzschild coordinates instead of the ECI coordinates that he used. (Also, he expands out the square roots and throws away higher order terms.) Note also that v is relative to a non-rotating observer, so an observer at rest on the Earth's equator has a v of about 450 m/s eastward.

So the ratio of clock rates for two observers, both at the same radius r but moving at different velocities, is:

\frac{d\tau_{1}}{d\tau_{2}} = \frac{\sqrt{1 - \frac{v_{1}^{2}}{c^{2}}}}{\sqrt{1 - \frac{v_{2}^{2}}{c^{2}}}}

where the terms in M cancel since both observers are at the same radius r. This does look different from the Doppler formula, but so what? Why would we expect the two to be the same for this case?

 

[PeterDonis]

Missed a couple of items in an earlier post, just wanted to clarify:

I think that you agree that
the instantaneous ratio (d\tau₁/dt)/(d\tau₂/dt) is always defined for any coordinate system, but is coordinate-dependent.

Yes.

Isn't "there is a timelike Killing vector field" and "there is a coordinate system in which the components of the metric are time-independent" the same thing?

Yes.

 

[Austin0]

Let me do an explicit calculation to prove my point.

In Rindler coordinates (X,T), we have two clocks, one at X = X₁, and one at X₂. The Rindler interval is:

d\tau² = X² dT² - dX²

So for clocks at rest in the X,T coordinates, we have:
d\tau = X dT

So the ratio of the rates is: d\tau₁/d\tau₂ = X₁/X₂

Conclusion: the "higher" clock (with greater X) runs faster.

Now, do the same calculation in the coordinate system (x,t) related to (X,T) through:

x = X cosh(gT)
t = X/c sinh(gT)

So d\tau² = dt² - 1/c² dx²
= dt² (1 - v²/c²)
where v = dx/dt = the speed of the clock. So

d\tau = \sqrt{1-(v/c)^{2}} dt


1)The ratios of the rates in this coordinate system is given by:
d\tau₁/d\tau₂ = \sqrt{1-(v_{1}/c)^{2}}/\sqrt{1-(v_{2}/c)^{2}}

At t=0, v_{1} = v_{2} = 0. So the ratio starts off equal to 1, not X₁/X₂.

x = X cosh(gT)
t = X/c sinh(gT)

So d\tau² = dt² - 1/c² dx²
= dt² (1 - v²/c²)
where v = dx/dt = the speed of the clock. So

d\tau = \sqrt{1-(v/c)^{2}} dt


The ratios of the rates in this coordinate system is given by:
d\tau₁/d\tau₂ = \sqrt{1-(v_{1}/c)^{2}}/\sqrt{1-(v_{2}/c)^{2}}


2)This part of the derivation is in error. If you did it correctly, you would have gotten that the correct result is \frac{d \tau_1}{d \tau_2}=\sqrt{\frac{1-v/c}{1+v/c}} where v is the instantaneous speed of the rocket containing the two clocks wrt the launcher frame.

Originally Posted by PeterDonis View Post

It looks to me like these two quantities refer to two different things. The first refers to Schwarzschild spacetime; the second refers to Rindler coordinates on Minkowski spacetime. The answers for those two cases will not be the same, because Schwarzschild spacetime is curved and Minkowski spacetime is flat.

No, both are about flat spacetime. The difference is that
√(1-(v₁/c)²)/√(1-(v₂/c)²) is the ratio of the two clock rates, as measured in the "launch" frame, while √(1-v/c)/√(1+v/c) is the redshift formula for the case in which the "lower" clock sends a signal while at rest, and the signal is received by the "upper" clock when that clock is traveling at speed v. (Since the light signal takes time to propagate, the upper clock will have achieved a nonzero velocity while the light signal is in flight).

3)My point is that the redshift formula is NOT the same as the ratio of clock rates, 4)except in very specific circumstances. Those circumstances actually hold for Rindler coordinates and for Schwarzschild coordinates, but they don't hold for arbitrary coordinates. The conditions for being able to equate "relative clock rates" with "redshift" are: (1) The metric tensor is independent of time, and (2) the sender and receiver are at rest in the coordinate system.

Perhaps you could clarify a point of confusion.
1)You stated that wrt the launch frame the front and rear clocks would be related by the ratio of gammas.

2)GAsahi declared that incorrect and said they would be related by the Doppler factor.
Indicating that they were not equivalent.

3)You then agreed with him that they were not equivalent in general

4) But were equivalent in this instance

So my question is:
Is the ratio of gammas as measured in the launch frame exactly the same ratio as derived from the Rindler metric ?

Ratio of Velocity gammas =Doppler ratio=Rindler ratio??

Or not?

thanks